Mixing
Volume of Intersection of cylinders (different radii)
$begingroup$
I want to derive a formula for the area of the intersection of two rigth cylinders with different radii. To get an idea I attached a sketch. 
My idea is to determine the borders of $x$ and $z$ in dependence of $y$.
The borders of $x$ are: $-sqrt{b^2-y^2} leq x leq sqrt{b^2-y^2} $
And the borders of z are $-sqrt{a^2-x^2}leq z leq sqrt{a^2-x^2}$
Thus, the resulting integral should be $int_{-b}^b int_{-sqrt{b^2-y^2}}^sqrt{b^2-y^2} int_{-sqrt{a^2-x^2}}^sqrt{a^2-x^2} ~mathrm{d}z~mathrm{d}x~mathrm{d}y$
Is this correct?
Thank you!
Fabian
calculus integration geometry volume faq
edited Jan 22 at 11:53
Lee David Chung Lin
4,37031242
asked Feb 27 '15 at 9:02
fmeyerfmeyer
1359
$endgroup$
add a comment |
$begingroup$
I want to derive a formula for the area of the intersection of two rigth cylinders with different radii. To get an idea I attached a sketch.
My idea is to determine the borders of $x$ and $z$ in dependence of $y$.
The borders of $x$ are: $-sqrt{b^2-y^2} leq x leq sqrt{b^2-y^2} $
And the borders of z are $-sqrt{a^2-x^2}leq z leq sqrt{a^2-x^2}$
Thus, the resulting integral should be $int_{-b}^b int_{-sqrt{b^2-y^2}}^sqrt{b^2-y^2} int_{-sqrt{a^2-x^2}}^sqrt{a^2-x^2} ~mathrm{d}z~mathrm{d}x~mathrm{d}y$
Is this correct?
Thank you!
Fabian
calculus integration geometry volume faq
edited Jan 22 at 11:53
Lee David Chung Lin
4,37031242
asked Feb 27 '15 at 9:02
fmeyerfmeyer
1359
$endgroup$
$begingroup$
This paper link (p.139) states that the volume is given by $int_{-b}^b int_{-sqrt{b^2-y^2}}^sqrt{b^2-y^2} int_{-sqrt{a^2-y^2}}^sqrt{a^2-y^2} ~mathrm{d}z~mathrm{d}x~mathrm{d}y$
$endgroup$
– fmeyer
Feb 27 '15 at 10:48
$begingroup$
This post is chosen to be the target/mother for (abstract) duplicates of the particular variation of orthogonal of different radii.
$endgroup$
– Lee David Chung Lin
Jan 22 at 11:55
add a comment |
$begingroup$
I want to derive a formula for the area of the intersection of two rigth cylinders with different radii. To get an idea I attached a sketch.
My idea is to determine the borders of $x$ and $z$ in dependence of $y$.
The borders of $x$ are: $-sqrt{b^2-y^2} leq x leq sqrt{b^2-y^2} $
And the borders of z are $-sqrt{a^2-x^2}leq z leq sqrt{a^2-x^2}$
Thus, the resulting integral should be $int_{-b}^b int_{-sqrt{b^2-y^2}}^sqrt{b^2-y^2} int_{-sqrt{a^2-x^2}}^sqrt{a^2-x^2} ~mathrm{d}z~mathrm{d}x~mathrm{d}y$
Is this correct?
Thank you!
Fabian
calculus integration geometry volume faq
edited Jan 22 at 11:53
Lee David Chung Lin
4,37031242
asked Feb 27 '15 at 9:02
fmeyerfmeyer
1359
$endgroup$
I want to derive a formula for the area of the intersection of two rigth cylinders with different radii. To get an idea I attached a sketch.
My idea is to determine the borders of $x$ and $z$ in dependence of $y$.
The borders of $x$ are: $-sqrt{b^2-y^2} leq x leq sqrt{b^2-y^2} $
And the borders of z are $-sqrt{a^2-x^2}leq z leq sqrt{a^2-x^2}$
Thus, the resulting integral should be $int_{-b}^b int_{-sqrt{b^2-y^2}}^sqrt{b^2-y^2} int_{-sqrt{a^2-x^2}}^sqrt{a^2-x^2} ~mathrm{d}z~mathrm{d}x~mathrm{d}y$
Is this correct?
Thank you!
Fabian
calculus integration geometry volume faq
calculus integration geometry volume faq
edited Jan 22 at 11:53
Lee David Chung Lin
4,37031242
asked Feb 27 '15 at 9:02
fmeyerfmeyer
1359
edited Jan 22 at 11:53
Lee David Chung Lin
4,37031242
asked Feb 27 '15 at 9:02
fmeyerfmeyer
1359
edited Jan 22 at 11:53
Lee David Chung Lin
4,37031242
edited Jan 22 at 11:53
Lee David Chung Lin
4,37031242
edited Jan 22 at 11:53
Lee David Chung Lin
4,37031242
4,37031242
asked Feb 27 '15 at 9:02
fmeyerfmeyer
1359
asked Feb 27 '15 at 9:02
fmeyerfmeyer
1359
asked Feb 27 '15 at 9:02
fmeyerfmeyer
1359
1359
$begingroup$
This paper link (p.139) states that the volume is given by $int_{-b}^b int_{-sqrt{b^2-y^2}}^sqrt{b^2-y^2} int_{-sqrt{a^2-y^2}}^sqrt{a^2-y^2} ~mathrm{d}z~mathrm{d}x~mathrm{d}y$
$endgroup$
– fmeyer
Feb 27 '15 at 10:48
$begingroup$
This post is chosen to be the target/mother for (abstract) duplicates of the particular variation of orthogonal of different radii.
$endgroup$
– Lee David Chung Lin
Jan 22 at 11:55
add a comment |
$begingroup$
This paper link (p.139) states that the volume is given by $int_{-b}^b int_{-sqrt{b^2-y^2}}^sqrt{b^2-y^2} int_{-sqrt{a^2-y^2}}^sqrt{a^2-y^2} ~mathrm{d}z~mathrm{d}x~mathrm{d}y$
$endgroup$
– fmeyer
Feb 27 '15 at 10:48
$begingroup$
This post is chosen to be the target/mother for (abstract) duplicates of the particular variation of orthogonal of different radii.
$endgroup$
– Lee David Chung Lin
Jan 22 at 11:55
$begingroup$
This paper link (p.139) states that the volume is given by $int_{-b}^b int_{-sqrt{b^2-y^2}}^sqrt{b^2-y^2} int_{-sqrt{a^2-y^2}}^sqrt{a^2-y^2} ~mathrm{d}z~mathrm{d}x~mathrm{d}y$
$endgroup$
– fmeyer
Feb 27 '15 at 10:48
$begingroup$
This paper link (p.139) states that the volume is given by $int_{-b}^b int_{-sqrt{b^2-y^2}}^sqrt{b^2-y^2} int_{-sqrt{a^2-y^2}}^sqrt{a^2-y^2} ~mathrm{d}z~mathrm{d}x~mathrm{d}y$
$endgroup$
– fmeyer
Feb 27 '15 at 10:48
$begingroup$
This post is chosen to be the target/mother for (abstract) duplicates of the particular variation of orthogonal of different radii.
$endgroup$
– Lee David Chung Lin
Jan 22 at 11:55
$begingroup$
This post is chosen to be the target/mother for (abstract) duplicates of the particular variation of orthogonal of different radii.
$endgroup$
– Lee David Chung Lin
Jan 22 at 11:55
add a comment |
1 Answer
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It is not as simple as you may think, and you shall run into trouble when coming to the second integral. The value of the innermost integral is not $2sqrt{a^2-x^2}$, it is $2sqrt{a^2-x^2}$ when $|x|leq a$ and $0$ when $xgeq a$. This has to be taken into account when doing the next integral, with respect to $x$. There will be cases to consider. Choosing another ordering of the variables, letting the variable $x$ being the outermost variable, will avoid running into cases. But wait.
Here is how I would go at it: I'm assuming $aleq b$. Looking at the figure we can immediately see that planes $x={rm const.}$ intersect the body $B$ whose volume we want to compute in rectangles $R_x$. It remains to find the area of $R_x$ and then to integrate over $x$. When $|x|>a$ then $R_x=emptyset$. When $|x|leq a$ then
$$R_x=bigl{(y,z)>bigm|>|y|leq sqrt{b^2-x^2}, |z|leqsqrt{a^2-x^2}bigr} .$$
It follows that
$${rm area}(R_x)=4sqrt{(b^2-x^2)(a^2-x^2)}qquad(-aleq xleq a) .$$
Therefore we obtain
$${rm vol}(B)=8int_0^a sqrt{(b^2-x^2)(a^2-x^2)}>dx .$$
I'm afraid that this is an elliptic integral when $b>a$.
answered Feb 27 '15 at 14:30
Christian BlatterChristian Blatter
175k8115327
$endgroup$
$begingroup$
You are absoulutely rigth. Thank you very much indeed.
$endgroup$
– fmeyer
Mar 1 '15 at 9:47
add a comment |
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1 Answer
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$begingroup$
It is not as simple as you may think, and you shall run into trouble when coming to the second integral. The value of the innermost integral is not $2sqrt{a^2-x^2}$, it is $2sqrt{a^2-x^2}$ when $|x|leq a$ and $0$ when $xgeq a$. This has to be taken into account when doing the next integral, with respect to $x$. There will be cases to consider. Choosing another ordering of the variables, letting the variable $x$ being the outermost variable, will avoid running into cases. But wait.
Here is how I would go at it: I'm assuming $aleq b$. Looking at the figure we can immediately see that planes $x={rm const.}$ intersect the body $B$ whose volume we want to compute in rectangles $R_x$. It remains to find the area of $R_x$ and then to integrate over $x$. When $|x|>a$ then $R_x=emptyset$. When $|x|leq a$ then
$$R_x=bigl{(y,z)>bigm|>|y|leq sqrt{b^2-x^2}, |z|leqsqrt{a^2-x^2}bigr} .$$
It follows that
$${rm area}(R_x)=4sqrt{(b^2-x^2)(a^2-x^2)}qquad(-aleq xleq a) .$$
Therefore we obtain
$${rm vol}(B)=8int_0^a sqrt{(b^2-x^2)(a^2-x^2)}>dx .$$
I'm afraid that this is an elliptic integral when $b>a$.
answered Feb 27 '15 at 14:30
Christian BlatterChristian Blatter
175k8115327
$endgroup$
$begingroup$
You are absoulutely rigth. Thank you very much indeed.
$endgroup$
– fmeyer
Mar 1 '15 at 9:47
add a comment |
$begingroup$
It is not as simple as you may think, and you shall run into trouble when coming to the second integral. The value of the innermost integral is not $2sqrt{a^2-x^2}$, it is $2sqrt{a^2-x^2}$ when $|x|leq a$ and $0$ when $xgeq a$. This has to be taken into account when doing the next integral, with respect to $x$. There will be cases to consider. Choosing another ordering of the variables, letting the variable $x$ being the outermost variable, will avoid running into cases. But wait.
Here is how I would go at it: I'm assuming $aleq b$. Looking at the figure we can immediately see that planes $x={rm const.}$ intersect the body $B$ whose volume we want to compute in rectangles $R_x$. It remains to find the area of $R_x$ and then to integrate over $x$. When $|x|>a$ then $R_x=emptyset$. When $|x|leq a$ then
$$R_x=bigl{(y,z)>bigm|>|y|leq sqrt{b^2-x^2}, |z|leqsqrt{a^2-x^2}bigr} .$$
It follows that
$${rm area}(R_x)=4sqrt{(b^2-x^2)(a^2-x^2)}qquad(-aleq xleq a) .$$
Therefore we obtain
$${rm vol}(B)=8int_0^a sqrt{(b^2-x^2)(a^2-x^2)}>dx .$$
I'm afraid that this is an elliptic integral when $b>a$.
answered Feb 27 '15 at 14:30
Christian BlatterChristian Blatter
175k8115327
$endgroup$
$begingroup$
You are absoulutely rigth. Thank you very much indeed.
$endgroup$
– fmeyer
Mar 1 '15 at 9:47
add a comment |
$begingroup$
It is not as simple as you may think, and you shall run into trouble when coming to the second integral. The value of the innermost integral is not $2sqrt{a^2-x^2}$, it is $2sqrt{a^2-x^2}$ when $|x|leq a$ and $0$ when $xgeq a$. This has to be taken into account when doing the next integral, with respect to $x$. There will be cases to consider. Choosing another ordering of the variables, letting the variable $x$ being the outermost variable, will avoid running into cases. But wait.
Here is how I would go at it: I'm assuming $aleq b$. Looking at the figure we can immediately see that planes $x={rm const.}$ intersect the body $B$ whose volume we want to compute in rectangles $R_x$. It remains to find the area of $R_x$ and then to integrate over $x$. When $|x|>a$ then $R_x=emptyset$. When $|x|leq a$ then
$$R_x=bigl{(y,z)>bigm|>|y|leq sqrt{b^2-x^2}, |z|leqsqrt{a^2-x^2}bigr} .$$
It follows that
$${rm area}(R_x)=4sqrt{(b^2-x^2)(a^2-x^2)}qquad(-aleq xleq a) .$$
Therefore we obtain
$${rm vol}(B)=8int_0^a sqrt{(b^2-x^2)(a^2-x^2)}>dx .$$
I'm afraid that this is an elliptic integral when $b>a$.
answered Feb 27 '15 at 14:30
Christian BlatterChristian Blatter
175k8115327
$endgroup$
It is not as simple as you may think, and you shall run into trouble when coming to the second integral. The value of the innermost integral is not $2sqrt{a^2-x^2}$, it is $2sqrt{a^2-x^2}$ when $|x|leq a$ and $0$ when $xgeq a$. This has to be taken into account when doing the next integral, with respect to $x$. There will be cases to consider. Choosing another ordering of the variables, letting the variable $x$ being the outermost variable, will avoid running into cases. But wait.
Here is how I would go at it: I'm assuming $aleq b$. Looking at the figure we can immediately see that planes $x={rm const.}$ intersect the body $B$ whose volume we want to compute in rectangles $R_x$. It remains to find the area of $R_x$ and then to integrate over $x$. When $|x|>a$ then $R_x=emptyset$. When $|x|leq a$ then
$$R_x=bigl{(y,z)>bigm|>|y|leq sqrt{b^2-x^2}, |z|leqsqrt{a^2-x^2}bigr} .$$
It follows that
$${rm area}(R_x)=4sqrt{(b^2-x^2)(a^2-x^2)}qquad(-aleq xleq a) .$$
Therefore we obtain
$${rm vol}(B)=8int_0^a sqrt{(b^2-x^2)(a^2-x^2)}>dx .$$
I'm afraid that this is an elliptic integral when $b>a$.
answered Feb 27 '15 at 14:30
Christian BlatterChristian Blatter
175k8115327
answered Feb 27 '15 at 14:30
Christian BlatterChristian Blatter
175k8115327
answered Feb 27 '15 at 14:30
Christian BlatterChristian Blatter
175k8115327
answered Feb 27 '15 at 14:30
Christian BlatterChristian Blatter
175k8115327
175k8115327
$begingroup$
You are absoulutely rigth. Thank you very much indeed.
$endgroup$
– fmeyer
Mar 1 '15 at 9:47
add a comment |
$begingroup$
You are absoulutely rigth. Thank you very much indeed.
$endgroup$
– fmeyer
Mar 1 '15 at 9:47
$begingroup$
You are absoulutely rigth. Thank you very much indeed.
$endgroup$
– fmeyer
Mar 1 '15 at 9:47
$begingroup$
You are absoulutely rigth. Thank you very much indeed.
$endgroup$
– fmeyer
Mar 1 '15 at 9:47
add a comment |
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$begingroup$
This paper link (p.139) states that the volume is given by $int_{-b}^b int_{-sqrt{b^2-y^2}}^sqrt{b^2-y^2} int_{-sqrt{a^2-y^2}}^sqrt{a^2-y^2} ~mathrm{d}z~mathrm{d}x~mathrm{d}y$
$endgroup$
– fmeyer
Feb 27 '15 at 10:48
$begingroup$
This post is chosen to be the target/mother for (abstract) duplicates of the particular variation of orthogonal of different radii.
$endgroup$
– Lee David Chung Lin
Jan 22 at 11:55